Physics 9702 · Paper 4

Photon Momentum

A photon has no mass, yet it carries momentum — and a stream of them presses on whatever they hit. A newer corner of the syllabus that Paper 4 now visits often.

p = E/c, and why it equals h/λ

p = \dfrac{E}{c} = \dfrac{h}{\lambda}

Symbols

$p$ = momentum of one photon (N s)
$E$ = photon energy (J)
$c$ = speed of light (m s⁻¹)
$\lambda$ = wavelength (m)

The two forms are the same fact: start from $p = E/c$ , substitute $E = hc/\lambda$ , and the $c$ cancels to give $p = h/\lambda$ — a “show that” in 2023 (9702/41/O/N/23 Q8(a)(i)).
When light hits a surface, the force is the rate of change of momentum of the arriving photons — exactly Newton's second law from AS. For full absorption each photon delivers $p$ , so $F = \dfrac{E}{c} \times \dfrac{N}{t} = \dfrac{P}{c}$ (9702/42/F/M/24 Q7(b)(ii)).
Reflection doubles it. A mirror sends each photon back the way it came, reversing its momentum, so the change is $2p$ and the force (and pressure) double (9702/41/O/N/23 Q8(b)(ii)).

Worked example

A half-and-half surface (2025 paper)

A 1.0 × 10−² W laser (photon energy $E$ , emitting $3.2\times10^{16}$ per second) hits a surface that absorbs half the photons and reflects the other half. Find the average force (9702/42/F/M/25 Q8(b)(iii)).

Absorbed photons each change momentum by $p$ ; reflected photons by $2p$ . The average per photon is $\tfrac{1}{2}(p + 2p) = \tfrac{3}{2}p$ .
$F = \dfrac{3}{2}\,\dfrac{E}{c}\,\dfrac{N}{t} = \dfrac{3}{2}\,\dfrac{P}{c} = \dfrac{3}{2}\,\dfrac{1.0\times10^{-2}}{3.00\times10^{8}} = 5.0\times10^{-11}\ \text{N}$ .

Answer

F = 5.0\times10^{-11}\ \text{N}

Common mistake

Same intensity, different colour, same pressure. Blue photons carry more momentum than red — but at the same intensity there are proportionally fewer per second, so the rate of momentum delivery, and hence the pressure, is unchanged (9702/41/O/N/23 Q8(c)).

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